The way to think about this is since S has no zero its minimal ideal inon-trivial. If it is not the whole semigroup you can factor out the minimal ideal to get a non trivial congruence.

So S must be its minimal ideal. An easy consequence of Rees's theorem is that Greens relations L and R are congruences on a finite simple semigroup. When you factor out L you get a right zero semigroup. Thus either S is a right zero semigroup or has a single L-class. Every equivalence relation on a right zero semigroup is a congruence. Since your semigroup has more than two elements it cannot be a right zero semigroup. So there is one L-class.

Dually there is one R-class. Thus S is a group and hence a simple group.

Now you should check if S has a zero and more than two elements then it must be a Rees Matrix semigroup over the trivial group with sandwich matrix containing no equal rows or columns and no zero rows or columns.

The way to think about this is since S has no zero its minimal ideal is non-trivial. If it is not the whole semigroup you can factor out the minimal ideal to get a non trivial congruence.

So $S$ must be its minimal ideal. An easy consequence of Rees's theorem is that Greens relations $\mathcal L$ and $\mathcal R$ are congruences on a finite simple semigroup. When you factor out $\mathcal L$ you get a right zero semigroup. Thus either $S$ is a right zero semigroup or has a single $\mathcal L$-class. Every equivalence relation on a right zero semigroup is a congruence. Since your semigroup has more than two elements it cannot be a right zero semigroup. So there is one $\mathcal L$-class.

Dually, there is one $\mathcal R$-class. Thus $S$ is a group and hence a simple group.

Now you should check if $S$ has a zero and more than two elements then it must be a Rees Matrix semigroup over the trivial group with sandwich matrix containing no equal rows or columns and no zero rows or columns.